im having problem with polynomial arithmetic?

Question

Compute the GCD of the following pairs of polynomials mod 7.
gcd (x3 + 6x2 + 6x + 5, x3 + x2 + 4x + 1)
Solution:
x + 5

i cant get the solution as above, plz help. tx.

Answer 1

Some of the other posters failed to notice that you
wanted the answer mod 7.
Let's divide each of your polynomials by x+5
but do all the arithmetic mod 7.
We get
x^3 + 6x² + 6x + 5 = (x+5)(x² + x + 1)
x^3 + x² + 4x + 1 = (x+5)(x² -4x +3),
all mod 7.
So x + 5 is certainly a common factor.
Let's show that the quotients have no
linear factor in common. Then we will be done!
We have
x² + x + 1 = (x-2)(x-4)
x² - 4x + 3 = (x-3)(x-1),
again, all mod 7.
So the quotients are relatively prime and the
GCD of your 2 original polynomials is indeed x + 5.

Answer 2

When you say mod7 do you mean the coefficients are the integers modulus 7? If you do then (using your notation)

x3 + 6x2 + 6x + 5 = (x + 5)(x2 + x + 1)
x3 + x2 + 4x + 1 = (x+5)(x2 + 3x + 3)

When doing the long division you have to rember that
-4 ≡ 3 (mod 7)
and
15 ≡ 1 (mod 7)

Answer 3

First, factor each of the two terms:

x^3 + 6x^2 + 6x + 5 is not too hard. Look at factors of 1 and 5, the two "end" coefficients, and they are 1 and 5. Also, since all the terms are positive, it is probably "(x+5) and not (x-5).
x+5 into the polynomial is x^2 + x + x
so the first polynomial can be rewritten as (x+5)*(x^2+x+1)
the second part of this does not factor because using the quadratic formula results in a square root -of 1*1-4*1*1 which is negative

So, let's now see if x+5 goes into the second polynomial. I can already see that this does not work, at least for whole numbers, since the last term, 1, is not divisible by 5. I would say that either the problem is written incorrectly, or the solution is wrong.

Answer 4

gcd (x3 + 6x2 + 6x + 5), (x3 + x2 + 4x + 1) =
gcd x(x3 + 6x2 + 6x + 5), x(x3 + x2 + 4x + 1) =
gcd x(x² + 6x + 6) + 5, x(x² + x + 4) + 1 = x + 5 and x + 1 =
gcd = x + 5
<><

Answer 5

You should do the Euclidean algorithm mod 7:

A: x^3+6x^2+6x+5 and B:x^3+x^2+4x+1
new A=B: x^3+x^2+4x+1 new B=A-B: 5x^2+2x+4
new A=B:5x^2+2x+4 new B=A-3x*B (mod 7!!!): 2x^2+6x+1
new A=B=2x^2+6x+1 new B=A+B: x+5
new A=B: x+5 new B=A-2x*B:3x+1

Note that 3(x+5) = 3x+15 = 3x=1 mod 7 so yes, (x+5) is g.c.d.

Answer 6

x^3+6x^2+6x+5
f(-5)=-125+150-30+5=0
so x+5 is a factor
dividing by x+5
1 6 6 5
0 -5 -5 -5
1 1 1 0
quotient is x^2+x+1
this is irreducible
x^3+x^2+4x+1
f(-5)=-125+25-20+1 is not zero
so x+5 cannot be the GCD
repost the sum

Answer 7

i'm with Chris in this one, i've got welded stainless for years. shop the warmth concentrated on the better ingredient of the steel, then path decrease back to the thinner cloth. I even have additionally stumbled on that a bite of distinctive cloth clamped (if that's a pipe it may be curved) below attracts maximum folk of the warmth faraway from the undertaking cloth. a distinctive style of steel like copper or aluminum, it wont merge along with your piece. It additionally enables an area the place you could upload cloth with out having it fall into the pipe. annoying slag skill the two no longer warm sufficient for separation of slag and steel or no longer waiting long sufficient for fabric to relax.

Answer 8

p(x) = 3x + 12x + 6x + 5
= 21x + 5
= 3(7)(x) + 5

q(x) = 3x + 2x + 4x + 1
= 9x + 1
= 3(3)(x) + 5

GCD = 3x + 5

hmmmmmmm..........
when you say 'x' in the question, do you mean a variable, 'x', or multiplication ?!

Answer 9

factor them mod 7 and look what they have in common.